2016 AMC 10A Problems/Problem 20

Problem

For some particular value of $N$ , when $(a+b+c+d+1)^N$ is expanded and like terms are combined, the resulting expression contains exactly $1001$ terms that include all four variables $a, b,c,$ and $d$ , each to some positive power. What is $N$ ?

$\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution 1

All the desired terms are in the form $a^xb^yc^zd^w1^t$ , where $x + y + z + w + t = N$ (the $1^t$ part is necessary to make stars and bars work better.) Since $x$ , $y$ , $z$ , and $w$ must be at least $1$ ( $t$ can be $0$ ), let $x' = x - 1$ , $y' = y - 1$ , $z' = z - 1$ , and $w' = w - 1$ , so $x' + y' + z' + w' + t = N - 4$ . Now, we use stars and bars (also known as ball and urn) to see that there are $\binom{(N-4)+4}{4}$ or $\binom{N}{4}$ solutions to this equation. We notice that $1001=7\cdot11\cdot13$ , which leads us to guess that $N$ is around these numbers. This suspicion proves to be correct, as we see that $\binom{14}{4} = 1001$ , giving us our answer of $N=\boxed{14}$ .

An alternative is to instead make the transformation $t'=t+1$ , so $x + y + z + w + t' = N + 1$ , and all variables are positive integers. The solution to this, by Stars and Bars is $\binom{(N+1)-1}{4}=\binom{N}{4}$ and we can proceed as above.

Solution 2

By Hockey Stick Identity, the number of terms that have all $a,b,c,d$ raised to a positive power is $\binom{N-1}{3}+\binom{N-2}{3}+\cdots + \binom{4}{3}+\binom{3}{3}=\binom{N}{4}$ . We now want to find some $N$ such that $\binom{N}{4} = 1001$ . As mentioned above, after noticing that $1001 = 7\cdot11\cdot13$ , and some trial and error, we find that $\binom{14}{4} = 1001$ , giving us our answer of $N=\boxed{14}$

Solution 3 (Stars and Bars)

5 sections ( $x,y,z,w,1$ ) 4 of which need at least 1 object. $\dbinom{N+4-4\cdot1}{4}$ . Test the choices and find that $N=\boxed{\text{(B) }14}$

Solution 4 (Casework)

The terms are in the form $a^xb^yc^zd^w1^t$ , where $x + y + z + w + t = N$ . The problem becomes distributing N identical balls to 5 different boxes $(x, y, z, w, t)$ with box $(x, y, z, w)$ with at least 1 ball. The $N$ balls in a row have

 gap between them.  We are going to put 4 or 3 divisors into those  gaps. There are  cases of how we count terms with .

Case $1$ : Count the terms that have at least $1$ of $(a, b, c, d, 1)$ . We put 4 divisors into $N-1$ gaps. There are $\binom{N-1}{4}$ terms.

Case $2$ : Count the terms that have at least $1$ of $(a, b, c, d)$ but not $1$ . We put 3 divisors into $N-1$ gaps. There are $\binom{N-1}{3}$ terms.

So, there are $\binom{N-1}{4}+\binom{N-1}{3}=\binom{N}{4}=1001$ terms. Similar to solutions $1, 2, 3$ , we find $N=\boxed{\text{(B) }14}$

~isabelchen

Video Solution

https://www.youtube.com/watch?v=R3eJW3PCYMs

Video Solution 2

https://youtu.be/TpG8wlj4eRA with 5 Stars and Bars examples preceding the solution. Time stamps in description to skip straight to solution.

~IceMatrix

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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